If f : ℝ → ℝ and g : ℝ → ℝ are two one to one functions, which of the following are
necessarily one to one on their domains? Explain why or why not by giving a proof or
an example.

f + g

fg

f3

f∕g

Draw the graph of the function f

(x)

= x3 + 1.

Draw the graph of the function f

(x)

= x2 + 2x + 2.

Draw the graph of the function f

(x)

=

x--
1+x

.

Suppose an =

1
n

and let nk = 2k. Find bk where bk = ank.

If Xi are sets and for some j, Xj = ∅, the empty set. Verify carefully that
∏i=1nXi = ∅.

Suppose f

(x)

+ f

(1)
x

= 7x and f is a function defined on ℝ∖

{0}

, the nonzero real
numbers. Find all values of x where f

(x )

= 1 if there are any. Does there exist any such
function?

Does there exist a function f, satisfying f

(x)

−f

(1-)
x

= 3x which has both x and

1
x

in
the domain of f?

In the situation of the Fibonacci sequence show that the formula for the nth term can
be found and is given by

-( -) -( -)
√ 5 1+ √ 5 n √ 5 1− √ 5 n
an = -5- --2--- − -5- --2--- .

Hint: You might be able to do this by induction but a better way would be to look
for a solution to the recurrence relation, an+2≡ an + an+1 of the form rn.
You will be able to show that there are two values of r which work, one of
which is r =

1+√5
2

. Next you can observe that if r1n and r2n both satisfy the
recurrence relation then so does cr1n + dr2n for any choice of constants c,d. Then
you try to pick c and d such that the conditions, a1 = 1 and a2 = 1 both
hold.

In an ordinary annuity, you make constant payments, P at the beginning of each
payment period. These accrue interest at the rate of r per payment period. This means
at the start of the first payment period, there is the payment P ≡ A1. Then this
produces an amount rP in interest so at the beginning of the second payment period,
you would have rP + P + P ≡ A2. Thus A2 = A1

(1 +r)

+ P. Then at the beginning of
the third payment period you would have A2

(1 + r)

+ P ≡ A3. Continuing in this way,
you see that the amount in at the beginning of the nth payment period would be An
given by An = An−1

(1+ r)

+ P and A1 = P. Thus A is a function defined on the
positive integers given recursively as just described and An is the amount at the
beginning of the nth payment period. Now if you wanted to find out An for large n, how
would you do it? One way would be to use the recurrance relation n times. A better way
would be to find a formula for An. Look for one in the form An = Czn + s
where C,z and s are to be determined. Show that C =

Pr-

,z =

(1+ r)

, and
s = −

Pr-

.

A well known puzzle consists of three pegs and several disks each of a different diameter,
each having a hole in the center which allows it to be slid down each of the pegs. These
disks are piled one on top of the other on one of the pegs, in order of decreasing
diameter, the larger disks always being below the smaller disks. The problem is to move
the whole pile of disks to another peg such that you never place a disk on a smaller disk.
If you have n disks, how many moves will it take? Of course this depends on n. If
n = 1, you can do it in one move. If n = 2, you would need 3. Let An be the
number required for n disks. Then in solving the puzzle, you must first obtain
the top n − 1 disks arranged in order on another peg before you can move
the bottom disk of the original pile. This takes An−1 moves. Explain why
An = 2An−1 + 1,A1 = 1 and give a formula for An. Look for one in the form
An = Crn + s. This puzzle is called the Tower of Hanoi. When you have found a
formula for An, explain why it is not possible to do this puzzle if n is very
large.